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48. (a) The nuclear reaction is written as
238
U
→
234
Th +
4
He
.
The energy released is
∆
E
1
=(
m
U
−
m
He
−
m
Th
)
c
2
=
(238
.
05079 u
−
4
.
00260 u
−
234
.
04363 u)(931
.
5MeV
/
u)
=4
.
25 MeV
.
(b) The reaction series consists of
238
U
→
237
U+
n,
followed by
237
U
→
236
Pa
+
p
236
→
235
Pa +
n
235
→
234
Th +
p
The net energy released is then
∆
E
2
m
238
U
−
m
237
U
−
m
n
)
c
2
+(
m
237
U
−
m
236
Pa
−
m
p
)
c
2
+(
m
236
Pa
−
m
235
Pa
−
m
n
)
c
2
m
235
Pa
−
m
234
Th
−
m
p
)
c
2
m
238
U
−
2
m
n
−
2
m
p
−
m
234
Th
)
c
2
=
[238
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 Fall '08
 SPRUNGER
 Physics, Energy

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